EEE225: Analogue and Digital Electronics - Lecture IX · 11/ 21 EEE225: Lecture 9 Opamps with...

EEE225: Analogue and Digital ElectronicsLecture IX

James E. Green

Department of Electronic EngineeringUniversity of Sheffield

[email protected]

1/ 21

2/ 21

EEE225: Lecture 9

This Lecture

1 Opamps with Frequency Dependent FeedbackPole-Zero CircuitsPassive and Active First Order Circuits: Standard FormsPassive and Active First Order Circuits: Low Pass with ‘k’Low Pass with ‘k’: Time and Frequency Domain ResponsePassive and Active First Order Circuits: High Pass with ‘k’High Pass with ‘k’: Time and Frequency Domain ResponsePole-Zero ResponsePassive PZ example: Getting the Standard Form...Active PZ example: Getting the Standard Form...

2 Review

3 Bear

3/ 21

EEE225: Lecture 9

Opamps with Frequency Dependent Feedback

Pole-Zero Circuits

Pole-Zero CircuitsPole-zero circuits aim to adjust the magnitude and phase responseof an analogue system. They are constructed from the standardamplifier blocks but with Z1 or Z2 having some frequencydependent components - almost always capacitors. Inductors aretoo imperfect1

−

+Av

Z1

ii

Z2 i f

vi vo

−

+

Z2

Z1

Av

vovi

1If an inductance is required, it may be manufactured with a capacitanceand an opamp or two forming a gyrator, a kind of impedance transformer. Seehttp://sound.westhost.com/articles/gyrator-filters.htm for examples.

4/ 21

EEE225: Lecture 9


Passive and Active First Order Circuits: Standard Forms

Standard FormsFirst order transfer functions fall into one of three standard forms,low pass,

vovi

= k1

1 + s τ= k

1

1 + j ωω0

= k1

1 + j ff0

(1)

high pass,

vovi

= ks τ

1 + s τ= k

j ωω0

1 + j ωω0

= kj ff0

1 + j ff0

(2)

and pole zero,

vovi

= k1 + s τ1

1 + s τ0= k

1 + j ωω1

1 + j ωω0

= k1 + j f

f1

1 + j ff0

(3)

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EEE225: Lecture 9


Passive and Active First Order Circuits: Low Pass with ‘k’

Passive and Active First Order: Low Pass with ‘k’

vi

R1

R2 C1 vo vi

R1−

+

C1

R2

vo∞

For the passive circuit:

R2

R1 + R2· 1

s C1 (R1//R2) + 1(4)

For the active circuit:

−R2

R1· 1

s C1 R2 + 1(5)

They are not identical! but they are similar in the shape of thefrequency response.

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EEE225: Lecture 9


Low Pass with ‘k’: Time and Frequency Domain Response

Time and Frequency Domain Response (Passive Version)

−60

−50

−40

−30

−20

−10

0

10−3 10−2 10−1 100 101 102 103

Normalized Frequency, ω [rads−1]

Vo

ViMagnitude[dBV]

−90

−75

−60

−45

−30

−15

0

Vo

ViPhase

[]

Magnitude [dBV]Phase []

00.10.20.30.40.50.60.70.80.91.01.1

0 1 2 3 4 5 6 7 8 9 10

Time Constants, τ

Norm

alisedVoltage[V

]

k

Vin

Vout

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EEE225: Lecture 9


Passive and Active First Order Circuits: High Pass with ‘k’

Passive and Active First Order: High Pass with ‘k’

vi

R1 C1

R2 vo vi

R1 C1−

+

R2

vo∞

For the passive circuit:

R2

R1 + R2· s C1 (R1 + R2)

s C1 (R1 + R2) + 1(6)

For the active circuit:

−R2

R1· s C1 R1

s C1 R1 + 1(7)

They are not identical! but they are similar in the shape of thefrequency response.

8/ 21

EEE225: Lecture 9


High Pass with ‘k’: Time and Frequency Domain Response

Time and Frequency Domain Response (Passive Version)

−60

−50

−40

−30

−20

−10

0

10−3 10−2 10−1 100 101 102 103

Normalized Frequency, ω [rads−1]

Vo

ViMagnitude[dBV]

0

15

30

45

60

75

90

Vo

ViPhase

[]


00.10.20.30.40.50.60.70.80.91.01.1

0 1 2 3 4 5 6 7 8 9 10

Time Constants, τ

Norm

alisedVoltage[V

]

k

Vin

Vout

9/ 21

EEE225: Lecture 9


Pole-Zero Response

Passive and Active First Order: Pole-Zero (or Zero-Pole)The PZ system is the linear sum of HP and LPThere is one pole and one zero.The pole may appear at a lower or higher frequency than thezero. The circuit is called pole-zero regardless!The pole determines the time constant, τOccasionally may be called lead or lag compensator in controlsystems discussion.

−60

−50

−40

−30

−20

10−3 10−2 10−1 100 101 102 103 104 105

Frequency, f [Hz]

Vo

ViMagnitude[dBV]

0

22.5

45.0

67.5

90.0

Vo

ViPhase

[]

ω1 ω0

20 dB/dec

kH

kL


10/ 21

EEE225: Lecture 9


Pole-Zero Response

There are two “gains” a low frequency (or DC, f → 0) gainand a high frequency (f →∞) gain, kL and kH respectively.

If zero frequency (ω1) < pole frequency (ω0) then kL < kHand phase “leads” (+ ve) between the pole and zero. This isthe case in the last slide.

If the zero frequency (ω1) > pole frequency (ω0) thenkL > kH and phase “lags” (- ve) between the pole and zero.

Magnitude slope tends to ±20 dB/dec as the system is firstorder. Phase tends to +90 or -90 depending on PZ or ZP butoften does not make it all the way.

Standard Forms:

frequency domain:

k1 + s τ1

1 + s τ0(8)

Alternatively:

k · 1

1 + s τ0+k · τ1

τ0· s τ0

1 + s τ0(9)

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EEE225: Lecture 9


Pole-Zero Response

The high frequency gain, kH = k · τ1

τ0and k = kL.

The step response depends on which of the pole or zero are at thelower frequency but for zero frequency < pole frequency we havesomething that is broadly HP looking but vout does not fall tozero, it tends towards kL. For zero frequency > pole frequency wehave something broadly LP but also having a finite kH .

00.10.20.30.40.50.60.70.80.91.01.1

0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0

Time Constants, τ

Norm

alisedVoltage[V

]

kL

kHVin

Vout

12/ 21

EEE225: Lecture 9


Passive PZ example: Getting the Standard Form...

Passive Pole-Zero ExampleFind the transfer function of the following PZ circuit.

vi

R3 C1

R1

R2 vo

Notice k is at the front and has noω dependence.

The s0 (unity) coefficient is 1 inthe numerator and denominator.

The highest power of s is one.

Always ask yourself, what is HFgain? what is LF gain? (goodsanity check)...

k · s τ1 + 1

s τ0 + 1=

R2

R1 + R2· s C1 (R1 + R3) + 1

s C1

(R2 R1 + R2 R3 + R1 R3

R1 + R2

)+ 1

(10)

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EEE225: Lecture 9



It’s a potential divider with R2 developing the output voltage,

vo =R2 vi

R2 + R1 //(R3 + 1

s C1

) (11)

Expanding,vovi

=R2

R2 +R1

(R3 + 1

s C1

)R1 + R3 + 1

s C1

(12)

Need to head towards 1 + s τ on the bottom. Multiply top(numerator) and bottom (denominator) by R1 + R3 + 1

s C1

R2

(R1 + R3 + 1

s C1

)R2

(R1 + R3 + 1

s C1

)+ R1

(R3 + 1

s C1

) (13)

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EEE225: Lecture 9



Multiplying out the brackets (expanding),

R2 R1 + R3 R2 + R2s C1

R2 R1 + R3 R2 + R2s C1

+ R1 R3 + R1s C1

(14)

Multiplying top and bottom by s C1,

(R2 R1 + R3 R2) s C1 + R2

s C1 R2 (R1 + R3) + R2 + R1 R3 s C1 + R1(15)

The unity term (coefficient of s0) in the denominator is R1 + R2.So lets divide top and bottom by R1 + R2 to get s τ + 1 on thebottom.

s C1R2(R1+R3)R1+R2

+ R2R1+R2

s C1(R2 R1+R2 R3+R1 R3)

R1+R2+ 1

(16)

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EEE225: Lecture 9



Having found the desired form of the denominator we know thepole has a time-constant, τ0 = C1

(R2 R1+R2 R3+R1 R3)R1+R2

. Thenumerator is still not in the right form though as it must be1 + s τ1. We need to divide the numerator by the numerator’spresent coefficients of s0, which are R2

R1+R2. We can’t change the

denominator though, it is already in the desired form, so we areunbalancing our expression. k, the frequency independent gain,will restore balance by becoming the unity coefficients of thenumerator, R2

R1+R2.

R2R1+R2

·

s C1 ·R2(R1+R3)

R1+R2

R2

R1+R2

+ R2

R1+R2

R2

R1+R2

s C1

(R2 R1+R2 R3+R1 R3)R1+R2

+ 1(17)

Performing the cancellations in (17) and bringing k outside of thefraction yields (10).

16/ 21

EEE225: Lecture 9


Active PZ example: Getting the Standard Form...

Active Pole-Zero Example

−

+

∞vi

R1

R3

C

R2

vo

Z1

Z2

HF gain: (at HF, C → 0 Ω)

vovi

=R2 + (R1//R3)

R1//R3(18)

LF gain: (at LF, C →∞ Ω)

vovi

=R2 + R1

R1(19)

This is a standard non-inverting amplifier which has the gainexpression:

vovi

=Z2 + Z1

Z1=

R2 + R1 //(R3 + 1

j ω C

)R1 //

(R3 + 1

j ω C

) (20)

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EEE225: Lecture 9



R2 +R1

(R3+ 1

j ω C

)R1+R3+ 1

j ω C

R1

(R3+ 1

j ω C

)R1+R3+ 1

j ω C

(21)

Multiply top and bottom by j ω C ,

R2 + R1 (R3 j ω C+1)1+j ω C(R1+R3)

R1 (R3 j ω C+1)1+j ω C(R1+R3)

(22)

Multiply top and bottom by 1 + j ω C (R1 + R3),

R2 (1 + j ω C (R1 + R3)) + R1 (1 + j ω C R3)

R1 (1 + j ω C R3)(23)

Collecting terms,

R1 + R2 + j ω (R2 R1 + R2 R3 + R1 R3) C

R1 (1 + j ω C R3)(24)

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EEE225: Lecture 9



Taking k outside, and comparing terms with the standard form,

R1 + R2

R1·

1 + j ω C(R2 R1+R2 R3+R1 R3

R1+R2

)1 + j ω C R3

≡ k1 + j ω τ1

1 + j ω τ0≡ k

1 + j ff11 + j ff0

(25)

f1 =R1 + R2

2π C (R1 R2 + R2 R3 + R1 R3)(26)

f0 =1

2π C R3(27)

k =R1 + R2

R1(28)

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EEE225: Lecture 9



when ω >> 2π f1 and 2π f0 (i.e. at high frequencies), the 1s arenegligible compared to the f terms,

∣∣∣∣vovi∣∣∣∣ = k

∣∣∣∣∣∣∣12 +

(ff1

)2

12 +(

ff0

)2

∣∣∣∣∣∣∣12

= kff1ff0

= kf0f1

(29)

kf0f1

= R1 + R2

R1·

12π C R3

R1+R22π C(R1 R2+R2 R3+R1 R3)

(30)

R1 R2 + R2 R3 + R1 R3

R1 R3=

R1 R2 + R2 R3

R1 R3+ 1 (31)

R2

(R1 + R2

R1 R3

)+ 1 =

R2

R1//R3+ 1 =

R2 + R1 //R3

R1 //R3(32)

Compare (32) with (18). At low frequencies,ω << 2π f1 and 2π f0, the 1s dominate the f terms, and gain→ k.

20/ 21

EEE225: Lecture 9

Review

Review

Revisited some EEE117 material on frequency and timedomain response of first order LP and HP systems.

Noted that the Pole-Zero circuit is a combination of the LPand HP first order circuits.

Enumerated some key points about the pole zerocircuit/system including:

There is one pole and one zeroThe pole can be found at a lower frequency than the zero orvice versa.The pole determines the time constant, τ .sometimes called “lead/lag compensation circuits”.

Examined a passive network pole zero circuit similar toEEE117

Examined an active, opamp based, pole zero circuit.

21/ 21

EEE225: Lecture 9

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EEE225: Analogue and Digital Electronics - Lecture IX · 11/ 21 EEE225: Lecture 9 Opamps with...

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